# Solve equation in `[0,2pi)` `1+sin(pi/2+x/2)=cos(21pi-x)`

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### 1 Answer

First, let us try to express the entire expression in terms of a single trigonometric function. We can do this by using trigonometric identities:

`cos(x) = sin( pi/2 - x)` [Complements]

`cos(x) = cos(-x)` [Even Functions]

`cos(x + 2pin) = cos(x)` [Periodicity]

`cos(x) = 2cos^2(x/2) - 1` [Half-Angle, variation]

`cos(x + pi) = -cos(x)` [Sum of Cosines]

Using all these, we can simplify the given expression:

`sin(pi/2 + x/2)` can be written as `cos(-x/2)` which can be written as `cos(x/2)` , by using the complement identity and the fact that cosine is an even function, respectively. On the other side, `cos(21pi - x)` can be written as `cos(20pi + (pi - x))` and since `20pi = 2*10pi` this is just equal to `cos(pi - x)` . Alternatively, using the last identity given above, this is just `-cos(-x)` or `-cos(x)` .` `

Hence, the entire expression can be re-written as:

`1+ cos(x/2) = -cos(x)`

Now, we solve for x.

We first transform this into a polynomial equation (on `cos(x/2)` ) using the identity derived from the half-angle identity given above:

`1 + cos(x/2) = -(2cos^2(x/2) - 1)`

`1 + cos(x/2) + 2cos^2(x/2) - 1 = 0`

`cos(x/2) + 2cos^2(x/2) = 0`

We can factor this as:

`cos(x/2) [ 1 + 2cos(x/2) ] = 0`

By the zero product property, we can split this into two equations:

`cos(x/2) = 0` and `1 + 2cos(x/2) = 0`

We now solve each equation.

From the first equation, we take `cos^-1` of both sides.

`cos^{-1}(cos(x/2)) = cos^{-1}(0)`

`x/2 = cos^{-1}(0) = pi/2 + pin, n in Z`

which is:

`x = pi + 2pin, n in Z`

Using the second equation:

`1 + 2cos(x/2) = 0`

`2cos(x/2) = -1`

`cos(x/2) = -1/2`

`cos^{-1}(cos(x/2)) = cos^{-1}(-1/2)`

`x/2 = cos^{-1}(-1/2)`

This is either:

`x/2 = (2pi)/3 + 2pin, n in Z` or `x/2 = (4pi)/3 + 2pi n, n in Z`

equivalent to:

`x = (4pi)/3 + 4pin, n in Z` or `x = (8pi)/3 + 4pin, n in Z`

Since we only want `x` such that `x in [0, 2pi)` we simply disregard the periodicity.

Hence,

`x = pi, (4pi)/3, (8pi)/3`